数据结构-队列应用(农夫过河问题)
1 // 数据结构 -队列运用(农夫过河问题)
2 //2009-6-23 by Larman Yuan C语言实现
3 //用四位二进制数分别顺序表示农夫、狼、白菜和羊
4 //用0表示在此岸 1表示在彼岸，那么初始状态为0000，终了状态为1111
5
6 #include "stdafx.h"
7 #include <stdlib.h>
8
9 #define MAXNUM 100
10
11 typedef int DataType;
12
13 struct SeqQueue
14 {
15 DataType q[MAXNUM];
16 int f,r;//f队头，r队尾
17 };
18 typedef struct SeqQueue * PSeqQueue;
19
20 PSeqQueue createEmptyQueue_seq(void)
21 {
22 PSeqQueue paqu;
23 paqu = (PSeqQueue)malloc(sizeof(struct SeqQueue));
24 if(paqu == NULL)
25 {
26 printf("Out of space!!\n");
27 }
28 else
29 {
30 paqu->f = 0;
31 paqu->r = 0;
32 }
33 return(paqu);
34 }
35
36 int isEmptyQueue_seq(PSeqQueue paqu)
37 {
38 return(paqu->f == paqu->r);
39 }
40
41 //是循环队列进队
42 void enQueue_seq(PSeqQueue paqu, DataType x)
43 {
44 if((paqu->r + 1) % MAXNUM == paqu->f)
45 printf("Full queue.\n");
46 else
47 {
48 paqu->q[paqu->r] = x;
49 paqu->r = (paqu->r + 1) % MAXNUM;
50 }
51 }
52
53 void deQueue_seq(PSeqQueue paqu)
54 {
55 if(paqu->f == paqu->r)
56 printf("Empty Queue.\n");
57 else
58 paqu->f = (paqu->f + 1) % MAXNUM;
59 }
60
61 DataType frontQueue_seq(PSeqQueue paqu)
62 {
63 return(paqu->q[paqu->f]);
64 }
65
66 int farmer(int location)
67 {
68 return (0 != (location & 0x08));
69 }
70
71 int wolf(int location)
72 {
73 return( 0 != ( location & 0x04));
74 }
75
76 int cabbage(int location)
77 {
78 return ( 0 != (location & 0x02));
79 }
80
81 int goat(int location)
82 {
83 return( 0 != (location & 0x01));
84 }
85
86 int safe(int location)
87 {
88 if((goat(location) == cabbage(location))&&
89 (goat(location) != farmer(location)))
90 return(0);
91 if((goat(location) == wolf(location)) &&
92 (goat(location) != farmer(location)))
93 return(0);
94 return(1);
95 }
96
97 void framerRiverProblem()
98 {
99 int movers, location, newLocation;
100 int route[16];//记录已考虑状态的路径
101 PSeqQueue moveTo;
102 moveTo = createEmptyQueue_seq();
103 enQueue_seq(moveTo,0x00);
104 for(int i = 0; i < 16; i++)
105 {
106 route[i] = -1;
107 }
108 route[0] = 0;
109 while(!isEmptyQueue_seq(moveTo) && (route[15] == -1))
110 {
111 location = frontQueue_seq(moveTo);
112 deQueue_seq(moveTo);
113 for(movers = 1; movers<= 8; movers <<= 1)//农夫总是在移动，随农夫移动的也只能是与农夫同侧的东西
114 {
115 if((0 != (location & 0x08)) == (0 != (location & movers)))
116 {
117 newLocation = location^(0x08 | movers);
118 if(safe(newLocation) &&(route[newLocation] == -1))
119 {
120 route[newLocation] = location;
121 enQueue_seq(moveTo, newLocation);
122 }
123 }
124 }
125
126 }
127 if(route[15] != -1)
128 {
129 printf("The reverse path is:\n");
130
131 for(location = 15; location >= 0; location = route[location])
132 {
133 printf("The location is: %d \n",location);
134 if(location == 0)
135 return;
136 }
137 }
138 else
139 {
140 printf("No solution.\n");
141 }
142 }
143 int _tmain(int argc, _TCHAR* argv[])
144 {
145 framerRiverProblem();
146 return 0;
147 }
148
2 //2009-6-23 by Larman Yuan C语言实现
3 //用四位二进制数分别顺序表示农夫、狼、白菜和羊
4 //用0表示在此岸 1表示在彼岸，那么初始状态为0000，终了状态为1111
5
6 #include "stdafx.h"
7 #include <stdlib.h>
8
9 #define MAXNUM 100
10
11 typedef int DataType;
12
13 struct SeqQueue
14 {
15 DataType q[MAXNUM];
16 int f,r;//f队头，r队尾
17 };
18 typedef struct SeqQueue * PSeqQueue;
19
20 PSeqQueue createEmptyQueue_seq(void)
21 {
22 PSeqQueue paqu;
23 paqu = (PSeqQueue)malloc(sizeof(struct SeqQueue));
24 if(paqu == NULL)
25 {
26 printf("Out of space!!\n");
27 }
28 else
29 {
30 paqu->f = 0;
31 paqu->r = 0;
32 }
33 return(paqu);
34 }
35
36 int isEmptyQueue_seq(PSeqQueue paqu)
37 {
38 return(paqu->f == paqu->r);
39 }
40
41 //是循环队列进队
42 void enQueue_seq(PSeqQueue paqu, DataType x)
43 {
44 if((paqu->r + 1) % MAXNUM == paqu->f)
45 printf("Full queue.\n");
46 else
47 {
48 paqu->q[paqu->r] = x;
49 paqu->r = (paqu->r + 1) % MAXNUM;
50 }
51 }
52
53 void deQueue_seq(PSeqQueue paqu)
54 {
55 if(paqu->f == paqu->r)
56 printf("Empty Queue.\n");
57 else
58 paqu->f = (paqu->f + 1) % MAXNUM;
59 }
60
61 DataType frontQueue_seq(PSeqQueue paqu)
62 {
63 return(paqu->q[paqu->f]);
64 }
65
66 int farmer(int location)
67 {
68 return (0 != (location & 0x08));
69 }
70
71 int wolf(int location)
72 {
73 return( 0 != ( location & 0x04));
74 }
75
76 int cabbage(int location)
77 {
78 return ( 0 != (location & 0x02));
79 }
80
81 int goat(int location)
82 {
83 return( 0 != (location & 0x01));
84 }
85
86 int safe(int location)
87 {
88 if((goat(location) == cabbage(location))&&
89 (goat(location) != farmer(location)))
90 return(0);
91 if((goat(location) == wolf(location)) &&
92 (goat(location) != farmer(location)))
93 return(0);
94 return(1);
95 }
96
97 void framerRiverProblem()
98 {
99 int movers, location, newLocation;
100 int route[16];//记录已考虑状态的路径
101 PSeqQueue moveTo;
102 moveTo = createEmptyQueue_seq();
103 enQueue_seq(moveTo,0x00);
104 for(int i = 0; i < 16; i++)
105 {
106 route[i] = -1;
107 }
108 route[0] = 0;
109 while(!isEmptyQueue_seq(moveTo) && (route[15] == -1))
110 {
111 location = frontQueue_seq(moveTo);
112 deQueue_seq(moveTo);
113 for(movers = 1; movers<= 8; movers <<= 1)//农夫总是在移动，随农夫移动的也只能是与农夫同侧的东西
114 {
115 if((0 != (location & 0x08)) == (0 != (location & movers)))
116 {
117 newLocation = location^(0x08 | movers);
118 if(safe(newLocation) &&(route[newLocation] == -1))
119 {
120 route[newLocation] = location;
121 enQueue_seq(moveTo, newLocation);
122 }
123 }
124 }
125
126 }
127 if(route[15] != -1)
128 {
129 printf("The reverse path is:\n");
130
131 for(location = 15; location >= 0; location = route[location])
132 {
133 printf("The location is: %d \n",location);
134 if(location == 0)
135 return;
136 }
137 }
138 else
139 {
140 printf("No solution.\n");
141 }
142 }
143 int _tmain(int argc, _TCHAR* argv[])
144 {
145 framerRiverProblem();
146 return 0;
147 }
148
运行结果：
The reverse path is:
The location is: 15
The location is: 6
The location is: 14
The location is: 2
The location is: 11
The location is: 1
The location is: 9
The location is: 0
